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On a scientific paper, I found the following equations about a compass gait (one leg behaves like an inverted pendulum, the other one as a simple pendulum; $\theta$ and $\phi$ are time-dependent):

$$ \ddot{\theta} - \sin\left( \theta - \gamma \right) = 0 $$

$$ \ddot{\theta} - \ddot{\phi} + \dot{\theta}^2 \, \sin(\phi) - \cos(\theta - \gamma) \, \sin(\phi) = 0 $$

I got the following ones, and also, in the paper, there are these before the sentence (*):

$$ \ddot{\theta} - \frac{g}{l} \, \sin\left( \theta - \gamma \right) = 0 $$

$$ \ddot{\theta} - \ddot{\phi} + \dot{\theta}^2 \, \sin(\phi) - \frac{g}{l} \, \cos(\theta - \gamma) \, \sin(\phi) = 0 $$

(*) The explanation of the paper to pass from the last two equations to the first two equations is "we have rescaled time by $\sqrt{l/g}$".

Could you tell me what it means?

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  • $\begingroup$ If you remove the last part "how to get the first two equations from the last two" then it will be on-topic. $\endgroup$ Commented May 15 at 14:25
  • $\begingroup$ Which scientific paper? Which page? $\endgroup$
    – Qmechanic
    Commented May 15 at 22:25
  • $\begingroup$ @Qmechanic pag. 2 of Mariano Garcia paper: The Simplest Walking Model: Stability, Complexity, and Scaling $\endgroup$ Commented May 16 at 8:51
  • $\begingroup$ Permalink: doi.org/10.1115/1.2798313 $\endgroup$
    – Qmechanic
    Commented May 16 at 8:55

2 Answers 2

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It's an usual procedure in deriving non-dimensional equations, from the dimensional ones: angles have no physical dimensions, they wanted a "scaled" (non-dimensional) time as well.

You just need to define a "scaled-time" independent variable

$$\tau = \sqrt{\frac{g}{\ell}} t $$

perform derivatives using the rules for composite functions,

$$\phi'(\tau) := \frac{d}{d\tau} \phi(t(\tau)) = \frac{dt}{d\tau} \frac{d}{dt} \phi(t) = \sqrt{\frac{\ell}{g}} \dfrac{d \phi}{d t} = \sqrt{\frac{\ell}{g}} \dot{\phi}(t) \ ,$$

$$\rightarrow \qquad \phi'' = \frac{\ell}{g} \ddot{\phi} \quad , \quad \phi'^2 = \frac{\ell}{g} \dot{\phi}^2$$

and multiply all the equations by the factor $\frac{g}{\ell}.$

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    $\begingroup$ You could add that if $g$ and $l$ mean what they usually mean, indeed $\sqrt{g/l}$ has dimension of $1/\mathrm{time}$, so $\tau$ is dimensionless $\endgroup$
    – Toffomat
    Commented May 15 at 10:08
  • $\begingroup$ Yeah, I was assuming that. Even if they don't mean what they usually mean, it's enough that $\theta$ is dimensionless (like angles are) and time $t$ has the proper dimension in the original equations: this is enough to imply that $\sqrt{g/l}$ has the physical dimension $1/\text{time}$ for the equations to be consistent $\endgroup$
    – basics
    Commented May 15 at 10:19
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    $\begingroup$ Let's go, anonymous downvoter! Keep pushing! $\endgroup$
    – basics
    Commented May 15 at 13:30
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It is the non-dimensionalization of the last two differential equations. Assuming $g$ as the acceleration due to gravity ($\text{m}/\text{s}^2$) and $l$ as the length (m), $\sqrt{l/g}$ has the dimension of time. Rescaling time by $\sqrt{l/g}$ means the substitution of the (physical quantity) time $t$ by its nondimensionalized counterpart $\tau=t/\sqrt{l/g}$.

Note that $\tau=t\sqrt{\frac gl}$ and $\frac{\rm d\tau}{{\rm d}t}=\sqrt{\frac gl}$.

I want to explain this in full detail since just writing $\theta(\tau)$, $\phi(\tau)$ and $\gamma(\tau)$ feels wrong (abuse of notation) and non-intuitive to me.

Let $\vartheta(\tau):=\theta(t)=\theta(\tau\sqrt{l/g})$, $\varphi(\tau):=\phi(t)=\phi(\tau\sqrt{l/g})$, and $\eta(\tau):=\gamma(t)=\gamma(\tau\sqrt{l/g})$.

Now, by the chain rule: $$\dot\theta:=\dot\theta(t)=\frac{{\rm d}\theta}{{\rm d}t} =\frac{{\rm d}\vartheta}{{\rm d}\tau} \frac{{\rm d}\tau}{{\rm d}t} =\sqrt{\frac{g}{l}} \ \dot \vartheta(\tau)$$ $$\ddot\theta:=\ddot\theta(t) =\frac{{\rm d}\dot\theta}{{\rm d}t} =\sqrt{\frac{g}{l}}\frac{{\rm d}\dot\vartheta}{{\rm d}\tau} \frac{{\rm d}\tau}{{\rm d}t} =\frac{g}{l}\ \ddot \vartheta(\tau)$$

Similarly, $$\dot\phi:=\dot\phi(t)=\sqrt{\frac{g}{l}} \ \dot \varphi(\tau),\ \ddot\phi:=\ddot\phi(t)=\frac{g}{l}\ \ddot \varphi(\tau)$$

Substituting these values into the last two differential equations, we get $$\color{red}{\boxed{\ddot{\theta} - \frac{g}{l} \, \sin\left( \theta - \gamma \right) = 0}}\implies\frac{g}{l}\ddot{\vartheta} - \frac{g}{l} \, \sin\left(\vartheta - \eta \right) = 0\implies\color{blue}{\boxed{\ddot{\vartheta} - \sin\left( \vartheta - \eta \right) = 0}}$$

$$\color{red}{\boxed{\ddot{\theta} - \ddot{\phi} + \dot{\theta}^2 \, \sin(\phi) - \frac{g}{l} \, \cos(\theta - \gamma) \, \sin(\phi) = 0}}\implies \frac{g}{l} \,\ddot{\vartheta} - \frac{g}{l} \,\ddot{\varphi} + \frac{g}{l} \,\dot{\vartheta}^2 \, \sin(\varphi) - \frac{g}{l} \,\cos(\vartheta - \eta) \, \sin(\varphi) = 0\implies\color{blue}{\boxed{\ddot{\vartheta} - \ddot{\varphi} + \dot{\vartheta}^2 \, \sin(\varphi) -\cos(\vartheta - \eta) \, \sin(\varphi) = 0}}$$ which are the required differential equations (in non-dimensionalized form) after rescaling time by $\sqrt{l/g}$.

I hope it helps!

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